1. The problem is to resolve the expression $$e^* = e_{inf} + \delta \epsilon^* \frac{1}{1 + i f / f_r}$$ into its real and imaginary components. 2. We start by identifying the terms: $$e_{inf}$$ is a real constant, $$\delta \epsilon^*$$ is a complex quantity, and $$i$$ is the imaginary unit. 3. The denominator is $$1 + i \frac{f}{f_r}$$. To separate real and imaginary parts, multiply numerator and denominator by the complex conjugate of the denominator: $$\frac{1}{1 + i \frac{f}{f_r}} \times \frac{1 - i \frac{f}{f_r}}{1 - i \frac{f}{f_r}} = \frac{1 - i \frac{f}{f_r}}{1 + \left(\frac{f}{f_r}\right)^2}$$ 4. Substitute back: $$e^* = e_{inf} + \delta \epsilon^* \frac{1 - i \frac{f}{f_r}}{1 + \left(\frac{f}{f_r}\right)^2}$$ 5. Let $$\delta \epsilon^* = \delta \epsilon' - i \delta \epsilon''$$ where $$\delta \epsilon'$$ and $$\delta \epsilon''$$ are the real and imaginary parts of $$\delta \epsilon^*$$. 6. Multiply: $$\delta \epsilon^* (1 - i \frac{f}{f_r}) = (\delta \epsilon' - i \delta \epsilon'')(1 - i \frac{f}{f_r}) = \delta \epsilon' - i \delta \epsilon' \frac{f}{f_r} - i \delta \epsilon'' + i^2 \delta \epsilon'' \frac{f}{f_r}$$ 7. Since $$i^2 = -1$$, simplify: $$= \delta \epsilon' - i \delta \epsilon' \frac{f}{f_r} - i \delta \epsilon'' - \delta \epsilon'' \frac{f}{f_r} = (\delta \epsilon' - \delta \epsilon'' \frac{f}{f_r}) - i (\delta \epsilon' \frac{f}{f_r} + \delta \epsilon'')$$ 8. Therefore, $$e^* = e_{inf} + \frac{(\delta \epsilon' - \delta \epsilon'' \frac{f}{f_r}) - i (\delta \epsilon' \frac{f}{f_r} + \delta \epsilon'')}{1 + \left(\frac{f}{f_r}\right)^2}$$ 9. The real part is: $$\text{Re}(e^*) = e_{inf} + \frac{\delta \epsilon' - \delta \epsilon'' \frac{f}{f_r}}{1 + \left(\frac{f}{f_r}\right)^2}$$ 10. The imaginary part is: $$\text{Im}(e^*) = - \frac{\delta \epsilon' \frac{f}{f_r} + \delta \epsilon''}{1 + \left(\frac{f}{f_r}\right)^2}$$ This completes the separation into real and imaginary components.